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We define $ (\mathbb{R}^m, \|.\|)$ to be a finite dimensional normed vector space with $ \|.\|$ is defined to be any norm in $ \mathbb{R^m}$. Let $S = \lbrace x \in \mathbb{R}^m: \| x\| = 1 \rbrace.$ Prove that $S$ is compact in $ (\mathbb{R}^m, \|.\|).$

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have you heard of Heine-Borel? – Holdsworth88 Mar 7 '12 at 8:16
@Sulaiman: Since you are happy with the answers for this questions, I suggest you accept one of them by ticking it. It makes this site much more organised. While I'm at it, I also suggest to you to accept the (good) answers to your other questions. – Michalis Mar 7 '12 at 19:09
@Michalis Thank you but I already did. In fact I ticked up both answers if this is not a problem. – Zizo Mar 7 '12 at 19:32
@Sulaiman: What you did is upvote. For the OP there is also another option, that marks a question as "answered". It is the tick right below the up/downvote buttons. It is important that you mark answered questions, as the users of this site will be able to concentrate on the unanswered ones. – Michalis Mar 7 '12 at 19:38
Thank you @Michalis! sure! done!! – Zizo Mar 7 '12 at 19:51
up vote 4 down vote accepted

You can use induction on $m$ and properties of $\mathbb R$ to show compacity using sequential compactness, which means the same thing for metric spaces. Now consider the norm induced on the space $\mathbb R \cong \mathbb R \times \{ 0 \} \times \dots \times \{ 0 \}$ viewed as a sub-metric-space of $\mathbb R^m$, and also consider the sequence $^1 x_n$ induced by putting all the other components but the first equal to $0$. Therefore the first component is a sequence of real numbers. Since in the reals, every metric is equivalent to the absolute value metric in the following sense $$ \forall (\mathbb R,d), \quad \exists c_1, c_2 > 0 \quad s.t. \quad \forall x,y \in \mathbb R, \quad c_1 d(x,y) \le |x-y| \le c_2 d(x,y). $$ One can deduce that the Bolzano-Weierstrass theorem also holds if we replace $| \cdot |$ by the induced metric from the norm in $\mathbb R^m$. Since the sequence $x_n$ is bounded, the sequence $^1x_n$ is also bounded in $\mathbb R$. Therefore there exists a subsequence of the sequence $x_n$ such that the first component converges. Repeat this procedure with the other components $^kx_n$ with $1 \le k \le n$, and you will get a subsequence that converges component by component, hence converges. This gives you for every sequence an element $x$ and a subsequence for which $x_n \to x$. Since $\| x_n \| = 1$ for every $n$, clearly $\|x \| = 1$, so that your subsequence converges in $S$ and we are done.

That is one way to do it ; if you have seen theorems in class that might help, perhaps they might make this less complicated.

Hope that helps,

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Thanks, this answered my question. Nice work. – Zizo Mar 7 '12 at 12:50
Is there any reference for this proof? – Zizo Mar 12 '12 at 3:30
I just wrote it ; so I guess you could call me the reference. – Patrick Da Silva Mar 12 '12 at 15:10

Use the Bolzano-Weierstrass theorem:

Since all the norms on $\mathbb{R}^m$ are equivalent, your subset will be closed and bounded in the euclidian norm $||\cdot||_2$, and hence compact.

Here is an exercise I found, that shows that all norms on $\mathbb{R}^m$ are equivalent:

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The compacity depends on the metric space, thus on the metric ; your way is one way to go, but it's not complete yet ; you need to show that compacity is a property invariant by equivalent metrics induced by norms. – Patrick Da Silva Mar 7 '12 at 8:56
The equivalence essentially relies on the fact that equivalent metrics induce open balls that can be included in one another. – Patrick Da Silva Mar 7 '12 at 9:17
@RagibZaman: This is not generally true, it depends on your base field. It is true if you have a complete valued field like $\mathbb{R},\mathbb{C}$ or $\mathbb{Q}_p$, but fails for example if you look at vector spaces over $\mathbb{Q}$ (e.g. number fields). – Michalis Mar 9 '12 at 11:21
@Patrick Da Silva: You are right, proving that the balls are included in each other you can show that two equivalent norms (with the "inequality"-definition) induce the same topology, now I understand your remark. – Michalis Mar 9 '12 at 11:25
@Michalis Sorry, I should have remembered that! Every normed vector space I've been studying lately has been over $\mathbb{R}$ or $\mathbb{C}$. – Ragib Zaman Mar 9 '12 at 13:31

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